New answers tagged

2

Hint: For a Hamiltonian $N$-body system with a phase space of dimension $2n=6N$ to be Liouville/complete integrable, there should exist $n$ constants of motion that are (i) independent and (ii) Poisson-commute, among other things. The 10 constants of motion mentioned on the Wikipedia page do not satisfy property (ii).


0

No it does not. This is because dS is a vector quantity whose direction,by convention, is radially outwards for a closed body (sphere,here). In this diagram,the field of q is towards left going towards the sphere. For the left half of the sphere the direction of dS is towards left.Roughly speaking, the field due to q and dS are parallel or their dot product ...


3

No, if Gauss' law gives $$\int_{\partial V}\vec E\cdot \text d\vec S=0 \tag{1}$$ this means that the total flux of the electric field through the surface is zero and it then follows that the total charge contained within that surface must also be zero. Note that in order for this integral to be zero, $\vec E$ doesn't have to be zero, what we're saying is ...


3

In the above types questions why we don't include the( initial and final both) velocities of bullet while finding the initial kinetic energy of the system in (COE) conservation of energy In the example you gave, kinetic energy is not conserved. During the completely inelastic collision some kinetic energy is spent on permanently deforming the pendulum and ...


0

The argument applies only in the centre of mass frame of the two pucks. I think this is the slick solution (otherwise you have to take into account the acceleration of the system to the right). I think it is not trivial to convince yourself that the argument is valid, because the centre of mass frame is not an inertial frame. It means introducing varying ...


0

(a) Suppose that just before the pucks collide the midpoint of the string has moved forward a distance $x$. The pucks have moved forward (that is in the $\vec F$ direction) by the smaller distance $(x-l)$. The forward component of the force that the string exerts on each puck is $F/2$ in magnitude, so the contribution of the acquired forward velocity to each ...


1

Let look at this example one dimensional Kinetic Energy is $$T=\frac{m}{2}\,\dot{x}^2$$ and Potential Energy is $$ U=U(x)$$ with Euler Lagrange you get $$m\,\ddot{x}+\frac{\partial}{\partial x}\,U(x)=0$$ thus according to Newton second law $$F=-\frac{\partial}{\partial x}\,U(x)$$ and the total energy $$E=T+U(x)=~\text{constant}$$ but if the potential energy ...


2

You already have proved the "only if" part. What your derivation shows is that $$\frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(t))\right)$$ equals zero only if $$\mathbf{F} = -\nabla V.$$ In other words, your work proves $$\mathbf{F} = -\nabla V \implies \frac{d}{d t}\left(\frac{1}{2} m|\dot{\mathbf{x}}(t)|^{2}+V(\mathbf{x}(...


1

Let me first discuss a reduced case, for the purpose of focus. (The concluding statement - all the way down - explains the non-equivalence.) I will first discuss the case of motion in a space with 1 spatial dimension. In classroom demonstrations this is of course the demonstration of two or more carts moving over an air track In this reduced environment, ...


1

The other answer is very good, but not useful if we don't understand Noether's theorem (or Lagrangians). Instead, we can reason about this in a less difficult but more roundabout way. You ask if Conservation of Energy is equivalent to Conservation of Momentum, i.e. $$CoE \iff CoP$$ First, we will disprove $$CoE \rightarrow CoP$$ We can do this by finding a ...


-1

For posterity I am adding another answer, showing specifically how to handle a single constrained rigid body (pinned) under the influence of an impulse along a specified direction. Consider a pined body with the center of mass at $\boldsymbol{c}$ and an impulse $J$ along the direction $\boldsymbol{n}$ at position $\boldsymbol{r}_{A}$. The pin is at the ...


4

is conservation of energy not equivalent to conservation of momentum? The conservation of energy is not equivalent to the conservation of momentum. Per Noetherโ€™s theorem momentum is conserved whenever the Lagrangian is symmetric under spatial translations and energy is conserved whenever the Lagrangian is symmetric under time translations. Since it is ...


1

Maybe it's useful to think this question from the Hamiltonian formalism. The Hamiltonian $\cal{H}$ of a system is, in general, the total energy of the system, expressed with a particular set of variables. For example, for a point particle moving in the presence of a constatn gravitational field, the energy is $$E = \frac{mv^2}{2} + mgx$$ but these variables ...


3

Noether's theorem is the thing to pay attention to here. This theorem basically says that if there is a symmetry in the underlying physics, there is also a conserved quantity. Traditionally the best-known symmetries are: Time invariance, i.e. the laws of physics do not change with time. If this is the case, then energy is conserved. Translational invariance,...


6

The question "what happens if a conservation law is observed to be broken?" is not a hypothetical one $-$ it is a historical one. The clearest example of this is the conservation of parity, a quantum-mechanical quantity which is related to the invariance of physical laws under spatial inversion. (There is a deep relation, called Noether's theorem, ...


2

My volume of Symon's Mechanics address' this question: The conservation laws are in a sense not laws at all, but postulates which we insist must hold in any physical theory. If, for example, for moving charged particles, we find that the total energy, defined as (T + V) [kinetic plus potential], is not constant, we do not abandon the law, but change its ...


1

the forces between two balls are always perpendicular to the surface, so in direction of the radius, where they touch. That is the reason you know the direction of the acceleration .


0

Now if we add the fluid, the fragments will slowly come to a halt. Assuming the same geometry and drag, if they slow down at the same rate, we can say that at any point in time ๐‘๐ด=โˆ’๐‘๐ต, and ๐‘๐‘ก๐‘œ๐‘ก๐‘Ž๐‘™=0 holds As far as I understand , momentum cannot be conserved if there is an external force on the system.Perhaps sir, what you are overlooking here is ...


0

Note: Your equation 2 has assumed $v_a'$ towards the right, and I will do the same for my equations. Also, I'm assuminng $d$ as the perpendicular distance between the centres of mass of the two rods (perpendicular to their lines of motion). $d=r+\frac{l}{2}cos(\alpha)$ $m_b v_a d = m_b d (v_a' - v_b') + I_b \omega_b' + I_a \omega_a'$ $m_a v_a d = m_a d (...


0

The fundamental tensor equation of relativistic mechanics of continous matter is $$ K^\mu = \partial_\mu T^{\mu\nu} $$ where $K^\mu$ is the 4-force-density acting on material medium and $T^{\mu\nu}$ is the energy-momentum-stress tensor of the system. Consistently for $\nu=0$ we obtain the equation of continuity and for $\nu=1,2,3$ the 3-vector equation of ...


0

The above formula is derived assuming external forces are present within the system. If there are no external forces $dp$ is equal to $0$. The mass accretion is similar to the process in a rocket system with direction of $u$ reversed. In free space there are no external forces on rocket therefore $F_{ext} = 0$. Similarly, near earth surface there exists ...


0

If we rewrite your expression (3) as ๐‘š๐‘‘๐ฏ + ๐ฏ๐‘‘๐‘š โˆ’๐ฎ๐‘‘๐‘š and set it to 0 (as it should be since there is no external force) we get d(m๐ฏ) = ๐ฎ๐‘‘๐‘š. So maybe all this is saying is that to this approximation of small changes, the change in the main object's momentum is given by the momentum of the small added mass.


1

In principle Noether's theorem (NT) itself also works for Galley's extended action functional (5) and its non-conservative potential $K$, which are still local in time. Galley's peculiar boundary conditions are not a problem as they are not relevant for NT. As usual it is your job to provide an off-shell quasisymmetry of the action in the first place. The ...


0

I'm assuming that the author of the problem is considering the gravitational force between the two spheres. Then you are looking for the momentum of a circular orbit.


2

First note that $$\nabla_\alpha(F_{\mu\nu}F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F_{\mu\nu}(\nabla_\alpha F^{\mu\nu})=F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})+F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})$$ $$=2F^{\mu\nu}(\nabla_\alpha F_{\mu\nu})=-2F^{\mu\nu}(\nabla _\mu F_{\nu\alpha} +\nabla _\nu F_{\alpha\mu}).$$ Using this equation, the anti-symmetry of $F_{...


0

A few notes first. I disagree with handling friction with a coefficient of restitution. It makes more sense to me with first calculating first the tangential impulse $J_e$ that would be needed for the parts not to slip past each other (in addition to the normal impulse $J_n$ due to bounce) and then capping the magnitude to a value such that $|J_e| \leq \mu | ...


2

Black hole will absorb the momentum and start to move with a constant speed. A black hole solution can be transformed to any other inertial frame via a global Lorentz transformation, so it can move with any speed through space.


0

Your premise for applying the law of conservation of angular momentum is wrong. The law can be applied if you find an axis about which the net torque on a given system is zero. In the case of a car taking a turn, the axis location changes from infinity to a closer point. The center of curvature of the car's path is not a valid axis as it changes with time. ...


0

Elementary particle physics view: The underlying level of nature is quantum mechanical and its behavior is encapsulated in the standard model of particle physics.. Charge is a conserved quantized quantity always carried by the charged particles in the table during interactions, and the probability of interaction is calculated using quantum field theory and ...


2

The entanglement is the coupling of different quantum systems. The experimenters bring entangled systems to their own journies. And when a quantum system is measured, the other quantum system is collapsed at the same time because they are entangled. While this process, no material is transferred. It is the non-locality of the world which makes this possible ...


3

Contrary to what science fiction may tell us, quantum teleportation does not involve the physical teleportation of matter. It only teleports the state of matter. In that sense, it teleports the information that is associated with a particle and not the particle itself. So, electric charge remains where it is. In response to the comment: the spin is a degree ...


0

Angular momentum is given by $mvr$. Angular momentum is a vector quantity. In this case angular momentum is not conserved in this case as direction of velocity is changing.


0

There are two aspects here: If you assuem a fixed (nonconstant) potential $V(x)$, then $H$ and $P$ don't commute. Also, momentum is not conserved, because particles are accelerated by the potential (e.g. are trapped in a potential well and bounce back and forth). Physically, this means that you considfer the potential as external to the system you're ...


2

If you just have the free particle Hamiltonian, with $V = 0$ then you'd expect that the dynamics are the same no matter the position of your system and that is precisely what happens. This corresponds, as you mentioned, to the fact that $\left[H,P\right]=0$. However, when you introduce a potential you break this translational symmetry because now the ...


1

The assertion/assumption Physics laws should not be changed under translations Does not imply the statement Hamiltonian must commute with the generator of translations If we have a system that has $[H,p]=0$ then itโ€™s a special system where the coordinates of the origin doesnโ€™t matter. Itโ€™s a spatially symmetric system. It has translational invariance. ...


4

For each continuity equation $$ \sum_{\mu = 0}^n\frac{\partial J^{\mu}_a}{\partial x^{\mu}}~=~0, \tag{A}$$ one can define a conserved quantity $$ Q_a(t)~:=~\int_V \! d^nx~ J^0_a(\vec{x}, t). \tag{B}$$ OP's example: Let $J^{\mu}=J^{\mu}_1 +i J^{\mu}_2$ be a complex current, and introduce a complex space coordinate $z=x^1+ix^2$, i.e. $n=2$. OP's continuity ...


0

Mass of ball, m = 0.2 Kg Acceleration due to gravity, g = 10m/sยฒ Heigh achieved by the ball, h = 2 m Length along which force is applied by hand, s = 0.2 m Force exerted by the hand, F is to be calculated With this data, we can calculate the force exerted by the hand using energy conservation, mgh = Fs (0.2)(10)(2) = F(0.2) 20 N= F Thus force exerted by the ...


2

The first thing you have to think is that it took energy to get 2m high, so it may be useful to calculate that energy. The final energy is given by: $mgh = 0.2 \cdot 10 \cdot 2 J = 4 J$ that energy is coming from the kinetic energy at the start of the motion, but that shange of speed is due to the force applied. Since the change of kinetic energy is equal to ...


1

After the comment by @ZeroTheHero I realized my answer is wrong. Here is the explanation of what went wrong. I will delete the answer later Conservation of ordinary momentum is implied by homogeneity of space and time, not just space. The most easiest way to see it is by the fact that you can simulate decelerating particle by adding a time dependent term to ...


-1

Let's say that anything moving relative to the "absolute rest" frame decelerates at a constant rate, $a = -x \ m/s^2$, until it is at rest with respect to the "absolute rest" frame. Then, if you are in the absolute rest frame, you will see this deceleration. If you are not in the absolute rest frame, but one that is moving at constant ...


2

For what it's worth, one may show that a 2nd-order ODE $$\ddot{x}~=~f(x,\dot{x})$$ without explicit time-dependence, or equivalently, a pair of 1st-order ODEs of the form $$ \dot{x}~=~v, \qquad \dot{v}~=~f(x,v),$$ always has a local Hamiltonian formulation, cf. e.g. this Phys.SE post. The Hamiltonian is a conserved quantity.


0

It is all essentially about the energy conservation. When forces do not depend on velocity, they are conservative, and characterized by the potential energy, so the total energy is conserved. When the forces do depend on the velocity, these may be either forces like Lorentz force that do not change the total energy or forces like viscous friction, when the ...


2

Charge conservation can be stated as a continuity equation $$\frac{\partial \rho}{\partial t}+\nabla\textbf{j}=0 \tag{1}$$ where $\rho$ is charge density (measurable in Coulomb/m$^3$) and $\textbf{j}$ is current density (measurable in Ampere/m$^2$). On the other hand, from the magnetostatic equation $$\nabla \times \textbf{B} = \mu_0\textbf{j} \tag{2}$$ you ...


4

Because Ping-Pong balls are elastic And in elastic collisions, kinetic energy is conserved. When your ball is allowed to bounce freely, it reaches a speed of zero at the top of its arc. At that point, all of its kinetic energy has been converted to gravitational potential energy. The opposite is true at the ground, where all its gravitational energy has ...


2

The quantity $C = g_{\alpha\beta}\dot{x}^\alpha K^\beta$ is a scalar invariant and is therefore independent of the coordinate system used. Since coordinate systems exist that are not singular at the event horizon (for example Kruskal Szekeres) you could calculate $C$ at the horizon using these coordinates. The quantity is not affected by a change of ...


0

There are several issues to discuss here. Since $\nabla\cdot J=\frac{\hbar}{2mi}(\Psi^\ast\nabla^2\Psi-\Psi\nabla^2\Psi^\ast)$, a TDSE solution satisfies$$\nabla^2\Psi=\frac{2m}{\hbar^2}(V\Psi-i\hbar\partial_t\Psi)\implies-\nabla\cdot J=\Psi^\ast\partial_t\Psi+\Psi\partial_t\Psi^\ast=\partial_t\rho,\,\rho:=\Psi^\ast\Psi.$$A choice of $V$ for which $\Psi:=\...


6

A solution of the free one-dimensional Schroedinger equation: $$ i\hbar \frac{\partial \psi}{\partial t} = -\frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2}\,\,\,\quad \text{(1)} $$ is: $$\psi = A e^{i(kx -\omega t)} \quad\quad\quad \text{(2)} $$ where $\omega$ fulfills the condition $\hbar \omega = \frac{(\hbar k)^2}{2m}$. If tentatively one ...


5

The continuity relation holds for solutions of the Schrodinger equation. $A\cos (\omega t - k x)$ is not a solution.


-1

To be precise the speed doesn't actually increase (actually it decreases as every bounce takes away some energy with it). Assuming it's an elastic collition, the speed remains constant, but the distance travelled decreases. As the ball travels shorter distance in shorter interval of time, it gives an illusion of increased speed.


69

Guy Inchbald mentions the slight force of the paddle on the ball and that velocity relative to the separation increases. For the latter, there's a further issue that every time the ball bounces, it makes a noise, and that noise becomes more frequent as the distance shortens, which increases the perception of speed. Also, I believe there is a third phenomenon:...


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